leetcode 70. Climbing Stairs QuestionEditorial

/*70. Climbing Stairs  QuestionEditorial Solution  My SubmissionsYou are climbing a stair case. It takes n steps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?解题思路：由于走到第n阶的最后一步，可以走一步或者两步，那么第n阶的步数等于第n-1阶步数+n-2阶步数。dp[n] = dp[n-1]+dp[n-2]*/#include <iostream>#include <string>#include <vector>using namespace std;class Solution {public:    int climbStairs(int n)     {        vector<int> dp(n+1, 0);        if (n < 3)            return n;        dp[0] = 0;        dp[1] = 1;        dp[2] = 2;        for (int i = 3; i <= n; ++i)            dp[i] = dp[i - 1] + dp[i - 2];        return dp[n];    }    int climbStairs1(int n)    {        if (n < 3)            return n;        int pre1 = 2;   //前一个的数字(n-1)的step        int pre2 = 1;   //n-2的step        int ret = 0;        int tmp = 0;        for (int i = 3; i <= n; ++i)        {            ret = pre1 + pre2;            //cout << ret << endl;            tmp = pre1; //更新pre1和pre2            pre1 = ret;            pre2 = tmp;        }        //cout << "---------------" << endl;        return ret;    }};void test(){    Solution sol;    cout << sol.climbStairs1(10) << endl;}int main(){    test();    return 0;}